Question

Four random numbers are chosen at random from {1,2,3,4…,40}. The probability that they are not consecutive.

Answer 1

Hint: This question requires the use of permutations, combinations and probability. Permutations is defined as the arrangement of r things out of n things whereas, combinations is defined as the selection of r things out of n things. Probability is a concept representing the possibility for the occurrence of any event.

Complete step-by-step answer:
Let’s first calculate the total number of cases, which can be obtained by selecting any four numbers from the given set of forty numbers.
\[ \Rightarrow T{ = ^{40}}{C_4} - - - - (i)\]
Now, for the favourable case lets first find its complement case, i.e., which is to select any four consecutive numbers from the given set of forty
\[ \Rightarrow {F^c} = 37 - - - - (ii)\]
Now, the probability of the event E can be given by
\[ \Rightarrow P(E) = 1 - P({E^c})\],where Ec represents the complement of the event E.
\[ \Rightarrow P(E) = 1 - \dfrac{{{F^c}}}{T}\]
\[ \Rightarrow P(E) = 1 - \dfrac{{37}}{{^{40}{C_4}}}\]
Now, the Value of \[^{40}{C_4}\]is 91390 and after 37 is divided by 91390 we get , \[\dfrac{1}{{2470}}\]
\[ \Rightarrow P(E) = 1 - \dfrac{1}{{2470}}\]
Now, subtracting the given fraction from 1 we get the final answer
\[ \Rightarrow P(E) = \dfrac{{2469}}{{2470}}\]
Thus, \[P(E) = \dfrac{{2469}}{{2470}}\] is the correct answer.

Note: The questions involve a lot of concepts like permutations, combinations and probability. One should be well versed with these topics before solving the question. One should be aware of the calculations and be sure of the correct answer.
Formulas used in this question are:
\[{ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{(n - r)!*r!}}\],where n=number of things and r=number of things taken certain time
\[{ \Rightarrow ^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}\], where n=number of things and r=number of things taken certain time
\[ \Rightarrow P(E) = \dfrac{F}{T},\] where F are the Favourable cases and T are the total number of cases
\[ \Rightarrow P(E) = 1 - P({E^c})\], where Ec is the complement of the given event E